Fig.1(Critically Damped System)
 Critically damped system(ξ=1): If the damping factor
ξ
is equal to one, or the damping coefficient c
is equal to critical damping coefficient "c_{c}", then the system is said to be a critically damped system.
$\xi=1 \quad \text OR \quad {c \over c_c} = 1\implies c = c_c$
 Two roots for critically damped system are given by S_{1} and S_{2} as below:
$S_1 = \big [\xi + \sqrt{\xi^2 1} \big] \omega_n \\
S_2 = \big [\xi  \sqrt{\xi^2 1} \big] \omega_n$
 For $ξ=1$; $S_1 = S_2 = \omega_n$
Here both the roots are real and equal, so the solution to the differential equation can be given by
$x = (A + Bt)e^{\omega_n t} \quad \quad ...(1)$
 Now differentiating equation (1) with respect to ‘t’, we get:
$\mathring x = Be^{\omega_nt}  \omega_n(A + Bt)e^{\omega_nt} \quad \quad ...(2)$

Now, let at
$t = 0$ : $x = X_0$
$t = 0$ : $\mathring x =0$

Substituting these values in equation (1):
$X_0 = A \quad ...(3)$
 Same way, from equation (2), we get
$\begin{aligned}
0&= B  \omega_n(A + 0) \\
0&= B  \omega_nA \\
B&= \omega_nA \\
B&= \omega_nX_0 \qquad ...(4)
\end{aligned}$
 Now putting the values of A and B in equation (1), we get:
$\begin{aligned}
x&= (X_0 + \omega_nX_0t)e^{\omega_nt} \\
x&= X_0(1 + \omega_nt)e^{\omega_nt} \qquad ...(5)
\end{aligned}$
Conclusion

From above equation (5), it is seen that as time t
increases, the displacement x
decreases exponentially.

The motion of a critically damped system is aperiodic (aperiodic motion motions are those motions in which the motion does not repeat after a regular interval of time i.e non periodic motion) and so the system does not shows vibrations.

For critically damped systems, if a system is displaced from its initial position, it will try to reach its mean position in a very short time.

Critically damped systems are generally seen in hydraulic doors closer as it is necessary for the door to come to its initial position in a very short time.
Fig.1(Critically Damped System)
 Critically damped system(ξ=1): If the damping factor
ξ
is equal to one, or the damping coefficient c
is equal to critical damping coefficient "c_{c}", then the system is said to be a critically damped system.
$\xi=1 \quad \text OR \quad {c \over c_c} = 1\implies c = c_c$
 Two roots for critically damped system are given by S_{1} and S_{2} as below:
$S_1 = \big [\xi + \sqrt{\xi^2 1} \big] \omega_n \\
S_2 = \big [\xi  \sqrt{\xi^2 1} \big] \omega_n$
 For $ξ=1$; $S_1 = S_2 = \omega_n$
Here both the roots are real and equal, so the solution to the differential equation can be given by
$x = (A + Bt)e^{\omega_n t} \quad \quad ...(1)$
 Now differentiating equation (1) with respect to ‘t’, we get:
$\mathring x = Be^{\omega_nt}  \omega_n(A + Bt)e^{\omega_nt} \quad \quad ...(2)$

Now, let at
$t = 0$ : $x = X_0$
$t = 0$ : $\mathring x =0$

Substituting these values in equation (1):
$X_0 = A \quad ...(3)$
 Same way, from equation (2), we get
$\begin{aligned}
0&= B  \omega_n(A + 0) \\
0&= B  \omega_nA \\
B&= \omega_nA \\
B&= \omega_nX_0 \qquad ...(4)
\end{aligned}$
 Now putting the values of A and B in equation (1), we get:
$\begin{aligned}
x&= (X_0 + \omega_nX_0t)e^{\omega_nt} \\
x&= X_0(1 + \omega_nt)e^{\omega_nt} \qquad ...(5)
\end{aligned}$
Conclusion

From above equation (5), it is seen that as time t
increases, the displacement x
decreases exponentially.

The motion of a critically damped system is aperiodic (aperiodic motion motions are those motions in which the motion does not repeat after a regular interval of time i.e non periodic motion) and so the system does not shows vibrations.

For critically damped systems, if a system is displaced from its initial position, it will try to reach its mean position in a very short time.

Critically damped systems are generally seen in hydraulic doors closer as it is necessary for the door to come to its initial position in a very short time.
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